Chapter 8: Chemical Composition

Question 1

atomic mass unit (amu)

Answer 1

A small unit of mass equal to 1.66 x 10²⁴ grams.

Question 2

average atomic mass

Answer 2

Average atomic mass = Σ (mass of isotope × relative abundance)

Ex: Average atomic mass of copper
= (62.93 amu × 0.6909) + (64.94 amu × 0.3091) = 63.55

Question 3

Calculating mass using atomic mass units

Answer 3

To solve this problem, use the average mass of the atom to set up an equivalence statement.

Ex: Calculate the mass, in amu, of a sample of aluminum that contains 75 atoms.

Equivalence statement:
1 Al atom = 26.98 amu

Conversion factor:
75 Al atoms x (26.98 amu/1 Al atom) = 2024 amu

Question 4

Calculating the number of atoms from the mass

Answer 4

To solve this problem, use the average atomic mass of the substance.

Ex: Calculate the number of sodium atoms present in a sample that has a mass of 1172.49 amu.

Equivalence statement:
1 Na atom = 22.99 amu

Conversion factor:

1172.49 amu x (1 Na atom/22.99 amu) = 51.00 Na atoms

Question 5

The Mole

Answer 5

The mole (mol) can be defined as the number equal to the number of carbon atoms in exactly 12.01 grams of pure ¹²C. Techniques for counting atoms very precisely have been used to determine this number to be

6.022 x 10²³

(This number is called Avogadro’s number.)

One mole of something consists of 6.022 x 10²³ units of that substance.

(In other words, one mole of anything is 6.022 x 10²³ units of that substance.)

Question 6

Avogadro’s Number

Answer 6

6.022 x 10²³

aka the mole (mol)

Question 7

A sample of any element that weighs ________________ to the _______________ of that element contains _______________ of that element.

Answer 7

A sample of any element that weighs a number of grams equal to the average atomic mass of that element contains 6.022 x 10²³ atoms (1 mole) of that element.

Question 8

A sample of an element with a mass ____________________ contains 1 mole of atoms.

Answer 8

A sample of an element with a mass equal to that element’s average atomic mass expressed in grams contains 1 mole of atoms.

Question 9

Calculating moles and number of atoms

Answer 9

Ex: Compute both the number of moles of atoms and the number of atoms in a 10.0-g sample of aluminum.

10.0 g Al ⇒ ? moles of Al atoms ⇒ number of Al atoms

1. Calculate the number of moles of aluminum atoms in a 10.0 g by using the equivalence statement:
   1 mol Al = 26.98 g Al
   to construct the appropriate conversion factor:
   10.0 g Al x (1 mol Al/26.98 g Al) = 0.371 mol Al
2. Next, convert from moles of atoms to the number of atoms, using the equivalence statment:
   6.022 x 10²³ Al atoms = 1 mol Al atoms
   to construct the appropriate conversion factor:
   0.371 mol Al x (6.022 x 10²³ Al atoms/1 mol Al) = 2.23 x 10²³​ Al atoms

Question 10

Molar Mass

Answer 10

The molar mass of any substance is the mass (in grams) of 1 mole of the substance. The molar mass is obtained by summin gthe masses of the component atoms.

Question 11

Calculating molar mass

Answer 11

Ex: Calculate the molar mass of sulfur dioxide, a gas produced when sulfur-containing fuels are burned.

Givens:

• The formula for sulfur dioxide is SO₂
• 1 mole of SO₂ molecules contains 1 mole of sulfur atoms and 2 moles of oxygen atoms:
  1 mol SO₂ = 1 mol S + 2 mol O
• Atomic masses of sulfur and oxygen:
  sulfur (32.07 g/mol), oxygen (16.00 g/mol)

Calculate:
Mass of 1 mol S = 1 x 32.07 g = 32.07 g
Mass of 2 mol O = 2 x 16.00 g = 32.00 g
Mass of 1 mol SO₂ = 64.07 g = molar mass

Answer:
The molar mass of SO₂ is 64.07 g. It represents the mass of 1 mole of SO₂ molecules.

Question 12

Formula Weight

Answer 12

A term sometimes used instead of molar mass for ionic compounds.

Question 13

Calculating mass from moles

Answer 13

Ex:

1) Calculate the molar mass of calcium carbonate.
2) A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this sample?

calcium carbonate = CaCO₃

1 mole of CaCO₃​ contains 1 mole of Ca, 1 mole of C, and 3 moles of O

Atomic masses:
calcium (40.08 g/mol)
carbon (12.01 g/mol)
oxygen (16.00 g/mol)

1. Calculate the molar mass of CaCO₃​:

mass of 1 mol Ca²⁺ = 1 x 40.08 g = 40.08 g
mass of 1 mol CO₃^2- (contains 1 mol C and 3 mol O)
1 mol C = 1 x 12.01 g = 12.01 g
3 mol O = 3 x 16.00 g = 48.00 g
mass of 1 mol CaCO₃​ = 100.09 g = molar mass
2. Calculate the mass of 4.86 moles of CaCO₃​:

4.86 mol CaCO₃​ x (100.09 g CaCO₃​/1 mol CaCO₃​) = 486 g CaCO₃

4.86 mol CaCO₃​ x conversion factor ⇒ 486 g CaCO₃​​

Question 14

Calculating moles from mass

Answer 14

Question 15

Calculating number of molecules

Answer 15

Ex:
Isopentyl acetate, C₇H₁₄O₂, the compound responsibile for the scent of bananas, can be produced commercially. Interestingly, bees release about 1 𝜇g (1 x 10^-6 g) of this compound when they sting. How many moles and how many molecules of isopentyl acetate are released in a typical bee sting?

isopentyl acetate = C₇H₁₄O₂

atomic masses:
carbon (12.01 g/mol)
hydrogen (1.008 g/mol)
oxygen (16.00 g/mol)

mass of C₇H₁₄O₂ = 1 x 10^-6 g

6.022 x 10²³ molecules in 1 mole

1. First compute the molar mass:
   7 mol C x 12.01 g/mol = 84.07 g C
   14 mol H x 1.008 g/mol = 14.11 g H
   2 mol O x 16.00 g/mol = 32.00 g O
   Molar mass = 130.18 g

This means that 1 mole of isopentyl acetate (6.022 x 10²³ molecules) has a mass of 130.18 g.
2. Determine the number of moles of isopentyl acetate in 1 𝜇g, which is 1 x 10^-6 g:

Equivalence statement:
1 mol isopentyl acetate = 130.18 g isopentyl acetate

Conversion factor:
1 x 10^-6 g C₇H₁₄O₂ x (1 mol C₇H₁₄O₂/130.18 g C₇H₁₄O₂)
= 8 x 10^-9 mol C₇H₁₄O₂
3. Determine the number of molecules:

​Equivalence statement:
1 mol = 6.022 x 10²³ units

Conversion factor:
8 x 10^-9 mol C₇H₁₄O₂ x (6.022 x 10²³ molecules/1 mol C₇H₁₄O₂)
= 5 x 10¹⁵ molecules

Question 16

Formula for Ethanol

Answer 16

C₂H₅OH

(NOT the expected formula C₂H₆O)

Question 17

Mass Percent

Answer 17

The percent by mass of a component of a mixture or of a given element in a compound.

Can be computed by comparing the mass of an element in 1 mole of the compound with the total mass of 1 mole of the compound, then multiplying the result by 100%.

Mass Percent = mass fraction x 100%

(Sometimes called the weight percent.)

Question 18

Mass Fraction

Answer 18

Mass fraction for a given element

=

mass of the element present in 1 mole of compound
mass of 1 mole of compound

Question 19

Empirical Formula

Answer 19

The empirical formula of a compound is the simplest whole-number ratio of the atoms present in the compound.

Question 20

The empirical formula can be found from _________________.

Answer 20

The empirical formula can be found from the percent composition of the compound.

Question 21

Molecular Formula

Answer 21

The molecular formula is the exact formula of the molecules present in a substance. It is always a whole-number multiple of the empirical formula.

Question 22

To calculate the empirical formula of a compound, we first determine ________________ of the various elements that are present.

Answer 22

To calculate the empirical formula of a compound, we first determine the relative masses of the various elements that are present.

Question 23

Steps for Determining the Empirical Formula of a Compound

Answer 23

1. Obtain the mass of each element present (in grams).
2. Determine the number of moles of each type of atom present.
3. Divide the number of moles of each element by the smallest number of moles (from step 2) to convert the smallest number to 1. If all of the numbers so obtained are integers, these are the subscripts in the empirical formula. If one or more of these numbers are not integers, go on to step 4.
4. Multiply the numbers derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula.

Question 24

Calculating Empirical Formulas from Percent Composition

Answer 24

Percent by mass for a given element means the grams of that element in 100 g of the compound.

Question 25

Knowing the composition of a compound in terms of the masses (or mass percentages) of the elements present, we can calculate the empirical formula but not the molecular formula. In order to obtain the molecular formula, we must know the ____________.

Answer 25

Knowing the composition of a compound in terms of the masses (or mass percentages) of the elements present, we can calculate the empirical formula but not the molecular formula. In order to obtain the molecular formula, we must know the molar mass.

Question 26

Calculating Molecular Formulas

Answer 26

Molecular formula = (empirical formula)_<em>n</em>

where n​ is a small whole number

• Thus, because:
  Molecular formula = n x empirical formula

…then:
Molar mass = n x empirical formula mass
* Solving for n gives:

n = molar mass
empirical formula mass

Question 27

The molar mass of any compound is the __________________ of ______________ of the compound.

Answer 27

The molar mass of any compound is the mass in grams of 1 mole of the compound.

Question 28

The molar mass of a compound is the _______________ of the __________________.

Answer 28

The molar mass of a compound is the sum of the masses of the component atoms.

Question 29

mass of a sample (g) = __________ x __________

Answer 29

mass of a sample (g) = moles of sample x molar mass of compound

Question 30

moles of a compound = __________ / __________

Answer 30

moles of a compound = mass of the same (g)
molar mass of the compound (g/mol)

Question 31

The empirical formula of a compound is the _________________.

Answer 31

The empirical formula of a compound is the simplest whole-number ratio of the atoms present in the compound.

Question 32

The empirical formula can be found from the ________________.

Answer 32

The empirical formula can be found from the percent composition of the compound.

Question 33

The molecular formula is the ______________________.

Answer 33

The molecular formula is the exact formula of the molecules present in a substance.

Question 34

The molecular formula is always __________________ of the empirical formula.

Answer 34

The molecular formula is always a whole-number multiple of the empirical formula.